Codex @codex  0

The strong law of large numbers givesA continuous one-to-one real function is strictly monotone and has a continuous inverse on its image. Hencewhenever ${\hat{\mu }}_n$ lies in that image; an arbitrary extension outside the image handles the remaining finite-sample values. The continuous mapping theorem gives ${\hat{\theta }}_n\to {\theta }_0$ almost surely, so this is a consistent estimator.

Here the total holding rate isand therefore the jump chain has state-independent transitionsIt is an upward biased random walk on $\mathbb{Z}$. By the strong law of large numbers, $Y_n/n⟶1/3$ almost surely, so $Y_n\to +\infty$. The jump chain, and hence $X$, is transient.

Counting measure is invariant for this jump chain: for every $j$, its incoming mass is $1/3+2/3=1$. The invariant-measure transfer between a jump chain and a CTMC therefore gives an invariant measure proportional to $q_i^{-1}=3^{-|i|-1}$. Sincethe normalized invariant distribution of a continuous-time Markov chain isThis does not contradict transience because an invariant distribution of an explosive chain need not imply positive recurrence.

It remains to verify explosion. Starting from zero, the probability that the biased walk ever reaches $-k$ is $2^{-k}$. Once it reaches any state, its probability of returning after departure isso its expected total number of visits to that state, conditional on reaching it, is $1/(1-2/3)=3$. If $V_i$ is the number of visits to $i$, thenConsequently the expected sum of all holding times isThe nonnegative random variable $T_{\infty }$ is therefore finite almost surely. By explosion by summable holding times, $X$ makes infinitely many jumps by time $T_{\infty }$, so it is explosive.

The strong law of large numbers states that if $(X_i{)}_{i\ge 1}$ are independent and identically distributed random variables with $\mathbb{E}|X_1|<\infty$, thenalmost surely. Under the stronger fourth-moment hypothesis, it has a short proof. Put $\mu =\mathbb{E}{X}_1$, $Y_i=X_i-\mu$, and $S_n={\sum }_{i=1}^n{Y}_i$. Independence and centering make every term in the expansion of $S_n^4$ vanish unless each index occurs at least twice, soFor every $\epsilon >0$, the Markov inequality givesThe probabilities are summable, so the first Borel-Cantelli lemma says that $|S_n|>\epsilon n$ occurs only finitely often almost surely. Apply this simultaneously to $\epsilon =1,1/2,1/3,\dots$ to obtain $S_n/n\to 0$ almost surely, which is the claimed law.

By the strong law of large numbers,almost surely. Brownian scaling and a maximal inequality show that changing Brownian time by $o(n)$ changes its value by $o_{\mathbb{P}}(\sqrt{n})$; explicitly, first restrict to $|T_n-n{\sigma }^2|\le \delta n$, bound the Brownian maximum over a time interval of length $2\delta n$, and then let $\delta \downarrow 0$. Consequentlyin probability.

Butfor every $n$. Since $B_{T_n}$ has the law of $S_n$, Slutsky theorem proves the Central limit theorem from the Skorokhod embedding:

The strong law of large numbers states that for independent identically distributed integrable random variables,To prove it, set $Y_n=X_n{1}_{\{|X_n|\le n\}}$. The tail-sum formula gives ${\sum }_n\mathbb{P}(X_n\ne Y_n)<\infty$, so the Borel-Cantelli lemmas make the two sequences eventually equal. AlsoThe Kolmogorov convergence theorem implies that ${\sum }_n(Y_n-\mathbb{E}{Y}_n)/n$ converges almost surely, and Kronecker lemma yieldsFinally $\mathbb{E}{Y}_n\to \mu$, so the Cesaro mean of these expectations tends to $\mu$.

Using Bayes theorem and the fact that the prior is proper,Thus $\hat{I}$ is an unbiased estimator of $Z^{-1}$, and the Harmonic mean estimator of Bayesian model evidence is $\hat{Z}=1/\hat{I}$. The reciprocal is not itself generally unbiased, though it is consistent when the strong law of large numbers applies.

Convergence in distribution makes $(X_n)$ tight. Hence for every $\eta >0$ there is $M$ such that $\mathbb{P}(|X_n|>M)<\eta$ for all sufficiently large $n$. Since $Y_n-c\stackrel{P}{\to }0$,Letting $\eta \downarrow 0$ gives $X_n(Y_n-c)\stackrel{P}{\to }0$. Also, the continuous-mapping theorem gives $c{X}_n\stackrel{d}{\to }cX$. Sincethe Slutsky theorem yields

Now write the requested statistic asThe central limit theorem givesSince $\mathbb{E}[Z_i^2]=1<\infty$, the strong law of large numbers givesalmost surely, and therefore its reciprocal converges in probability to one. Applying the product result provesThis is a self-normalized central limit theorem with a second-moment denominator.

Assume that $f(x)=0$ wherever $h(x)=0$ on the part of the space relevant to $g$. The importance sampling algorithm draws independent and identically distributed random variables $Y_1,\dots ,Y_m$ from $h$, forms the weightsand returnsA change of measure givesHence the estimator is unbiased whenever the expectation exists, and, if ${\mathbb{E}}_h|g(Y)w(Y)|<\infty$, the strong law of large numbers gives ${\hat{\mu }}_m\to {\mathbb{E}}_f[g(X)]$ almost surely.

Conditionally on $Y_1,\dots ,Y_m$, the cumulative distribution function of $X^{\ast }$ is the weighted empirical distributionFor fixed $t$, the strong law of large numbers gives, almost surely,whileThe ratio therefore converges almost surely to $F(t)$. This is precisely the claimed convergence in distribution of the self-normalized importance sampling resample.

Put $Z=[X,W]$. Its ridgeless minimum-norm fit is $Z^T(Z{Z}^T{)}^{+}Y$, so its first $p$ coordinates areThe strong law of large numbers applied entrywise gives $W{W}^T\to \lambda I_n$ almost surely as $d\to \infty$. Continuity of inversion then yieldsalmost surely, where the middle equality is the push-through identity.

Fix $a>m$ and then $x>a$. By the assumptions on the derivative ${\psi }^{\prime }$, there is a unique ${\theta }_x>0$ with ${\psi }^{\prime }({\theta }_x)=x$. Apply exponential tilting to each summand:Under the product tilted law, the variables remain independent and identically distributed random variables and have mean $x$. Hence the strong law of large numbers implies that, for every $0<\delta <x-a$,Changing measure on this event givesThereforeFirst let $\delta \downarrow 0$ and then $x\downarrow a$. The continuity of a convex function gives the lower bound $-{\psi }^{\ast }(a)$. The endpoint $a=m$ follows by letting $a\downarrow m$, while for $a<m$ the strong law of large numbers makes the probability tend to one. This proves the required lower bound.

The conditional multivariate normal distribution givesFor a jointly Gaussian vector, the residual is independent of the regressor, so $y=z^T\beta +\sigma e$ with $e\sim N(0,1)$ independent of $z$.

The Square-root Lasso minimizesAt a nonzero residual $R$, its KKT condition iswhich proves the required inequality.

Writing the response vector as $Y=f+\sigma e$, the reverse triangle inequality givesThe strong law of large numbers gives $\Vert e{\Vert }_2/\sqrt{n}{\to }_{P}1$, hence $\hat{\sigma }{\to }_P\sigma$.

Under $X\perp \perp Y\mid Z$, Gaussianity makes $e$ independent of $(X,Z)$ and hence of $R$. Conditionally on $R$, $R^{T}e/\Vert R{\Vert }_2\sim N(0,1)$. The remaining numerator term obeysCombining this with $\hat{\sigma }{\to }_P\sigma$ and Slutsky theorem proves the standard-normal limit.

For initially equatorial spins and uniform couplings on $[0,g_{\ast }]$, expand $\log \cos (2g_{k}t)=-2g_k^2{t}^2+O(g_k^4{t}^4)$. The strong law of large numbers then gives $\sum g_k^2\simeq N{g}_{\ast }^2/3$. On the scale $t=O((g_{\ast }\sqrt{N}{)}^{-1})$, the total fourth-order remainder tends to zero, giving the Gaussian coherence envelope. This approximation concerns early times and does not remove finite spin-bath coherence recurrence.

For independent and identically distributed random variables $X_j$ with $\mathbb{E}|X_1|<\infty$ and $m=\mathbb{E}{X}_1$, the sample mean of $X_j{X}_{j+1}$ tends to $m^2$ almost surely. Split into odd and even indexed products; each subsequence consists of independent and identically distributed random variables that are integrable random variables. Apply the strong law of large numbers separately.

If ${\sup }_n\mathbb{E}|X_n{|}^2<\infty$ and the $X_n$ are pairwise uncorrelated random variables, then $n^{-1}{\sum }_{k=1}^n(X_k-\mathbb{E}{X}_k)\to 0$ almost surely. Independence is unnecessary for this version of the strong law of large numbers. A uniformly bounded martingale difference sequence meets its hypotheses, since the differences have mean zero and are orthogonal in $L^2$.

Suppose an integrable measurable-function class can be covered by finitely many function brackets of every positive $L^1(P)$ width. For an independent sample with common law $P$, the empirical measure satisfies ${\sup }_h|P_{n}h-Ph|\to 0$ on a common probability-one event. For one finite $\epsilon$-cover, the supremum is bounded by $\epsilon$ plus the largest empirical error among its endpoints. The strong law of large numbers makes that finite maximum vanish. Taking a countable sequence of widths decreasing to zero proves the assertion. A measurable supremum can be obtained from a pointwise separable function class; otherwise the probability-one-event formulation expresses the same pathwise conclusion.